Formally, the weak topology on some locally convex space $X$ can be defined as the Initial Topology with respect to the topological dual $X^*$, i.e. the weakest topology that makes all $f\in X^*$ continuous. In this sense, your first statement
Can we say that $f$ is weakly continuous if $f^{−1}(B)$ is weakly open, i.e., $f^{−1}(B)$ belongs to weak topology for every $B$ being open in $R$?
and your third one
But also, every $f^{−1}_j(B_j)$ belongs to the base of the weak topology, which automatically makes them weakly continuous.
are correct. Your second statement however
Since weak topology is weaker than the normed one, i.e., every weakly open set is strongly open, then the above definition may have sense - weak continuity is something more genreal then continuity. Am I correct?
is false, assuming that "more general" means that the class of weakly continuous functions is contained in the continuous functions - it is exactly the opposite way around: Every weakly continuous function is continuous, but there are continuous functions that are not weakly continuous. A classic example is the function $x\mapsto\|x\| $ on infinite-dimensional Hilbert Spaces - "normal" continuity is trivial, but this function is not weakly continuous since an orthonormal set weakly converges to zero, but has constant $1$ as image. I like to think of this in the sense of sequential continuity: In the weak topology, there are "more" converging sequences than in the standard topology, hence the requirement
$$
\lim_n f(x_n)=f(\lim_n x_n)
$$
for all convergent sequences $(x_n)_{n}$ is harder to fulfil. In general, I find that thinking about the different topologies with respect to the sequences rather than the open sets is often very helpful, although you should be careful with that when it comes to proofs: The weak topology is in general not first-countable.
One goes about understanding a topology by looking at the basis, the "simplest" open sets of which all other sets are composed as unions. For a metric space, the basis consists of the balls, of course. And what's even nicer about topological vector spaces is that you only have to study what the basis sets look like at the origin, since topological vector spaces are shift-invariant.
The norm-topology's basis at the origin consists of the balls $ B(0,r) $. Easy enough. What do basis sets look like in the weak topology? Take a bounded linear functional $ f \in X^* $, and then look at $ \{ x \in X : |f(x)| < r \} $. You can hopefully imagine, this is "the space that is sandwiched between two hyperplanes". Now, this isn't quite the basis. To get a basis (at the origin), you need to consider all possible intersections of these "sandwiches". But the important thing is, you only get to "sandwich" in a finite number of directions $ f_1, \ldots, f_N $. So, in the infinite dimensional case, there's always some direction that you fail to sandwich. Hence, these basis sets are not bounded. This is counterintuitive, because in $ \mathbb{R}^3 $ (for example), you obtain a bounded set once you "sandwich" along the x-axis, y-axis, and z-axis (or in 3 directions of your choice).
Question 1 Let's first get clear on why the norm-topology contains the weak topology. Let $ U $ be open in the weak topology. Fix a point $ x_0 \in U $. We can find a basis set $ V $ with $ x \in V \subseteq U $. Recall, by "basis set", I mean a finite number of sandwiches, but we can be rigorous about this: for some numbers $ m_1 < M_1, \ldots m_N < M_N $ and $ f_1, \ldots f_N \in X^* $, the basis set $ V $ can be expressed
$$
V = \{ x \in X : m_1 < f_1(x) < M_1, \ldots, m_N < f_N(x) < M_N \}
$$
Now, it's pretty straightforward to find a ball $ B(x_0, r_0) $ that fits inside of $ V $. Do you see how? It relies on the boundedness of hte functionals $ f_1, \ldots f_N $. Thus we have
$$
x_0 \in B(x_0, r_0) \subseteq V \subseteq U
$$
which, you'll recall, shows that $ U $ is open in the norm topology (make sure you see why).
Question 1, Part b You then asked why $ x \mapsto |f(x)| $ is norm-continuous. But this is a composition of two continuous functions, the absolute value function and $ f \in X^* $ (which is continuous by definition).
Question 2 You asked about showing that $ g : x \mapsto \|Lx\| $ is weakly continuous, where $ L:X \rightarrow X $ is a bounded linear map. Surprisingly, it is not continuous with respect to the weak topology (in general). Indeed, pick $ X $ to be your favorite infinite-dimensional space, let's say $ X = \mathcal{l}^1(\mathbb{N}) $ and $ L = Identity : X \rightarrow X $. Then I claim $ g : x \mapsto \|Lx\| $ is not weakly continuous, since indeed $ g^{-1}((-a,a)) = B(0,a) $ is bounded, whereas weakly open sets in $ \mathcal{l}^1 $ are not bounded (!).
Addendum Fundamentally, the thing that's surprising here is that the norm $ x \mapsto \|x\| $ is not weakly continuous. To try to wrap our heads around this, consider the weak basis we discussed above, composed of finite intersections of "hyperplane sandwiches." This basis of the weak topology tells us that, in order for a function $ f : X \rightarrow \mathbb{R} $ to be weakly continuous, it is only allowed to change in a finite number of directions. The norm function, on the other hand, changes in all directions that you walk away from the origin.
Best Answer
First a few comments about your statement:
1) I assume you are thinking in a certain $f$ which is a continuous linear functional (otherwise the argument does not say anything).
2) What do you mean by "a basis element of the the vector space". If you mean any element $e$ such that $\{e\}$ can be extended to a basis of the space, then $e$ can be any non $0$ element (which is really not saying much).
3) Note that you are not really using the finite dimensional structure, what hints that there must be something wrong in the argument since the result does not hold for infinite dimensional spaces.
Indeed, what fails in your argument is that it is not true in general that you can find $(u,v)$ such that $f^{-1}(u,v)$ is included in $B(a, \epsilon)$. This is the case in any infinite dimensional normed space, since it can be proved that the non empty weak open sets always contain a line.
Now, a way to prove the result:
I want to show that given a sequence $(x_n)\subset X$ and $x\in X$ such that $x_n\to x$ weakly, then $x_n\to x$ in the norm topology.
Let $\{e_1,\dots,e_n \}$ be a basis of the vector space and denote by $\{e_1^*,\dots,e_n^* \}$ its dual basis. The convergence of the sequence in weak topology says in particular that $e^*_i(x_n)\to e^*_i(x)$ for all $i$. Then:
$$x_n= \sum_{i=1}^n e^*_i(x_n)\to \sum_{i=1}^n e^*_i(x) = x,$$
where the last convergence is in the norm topology.